Partial Derivatives used in a manufacturing scenario word problem

In summary, the conversation discusses finding partial derivatives and their significance in the context of manufacturing blackboards and whiteboards. The partial derivatives are calculated using Wolfram Alpha and the resulting answer for dp/dy(50,25) is a negative value, indicating a monthly loss of $13.93 when selling 50 blackboards and 25 whiteboards.
  • #1
Meow101
2
0

Homework Statement


A) if [PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP10819hf7feh5hf5hh7e00004ga728h1d42i6h0c?MSPStoreType=image/gif&s=18&w=161&h=20
Find the partial derivatives?

B) Suppose you are manufacturing blackboards and whiteboards and that P is your monthly profit when your selling price for a blackboard is $x and for a whiteboard is $y. compute dp/dy(50,25) and explain what your answer means in the context of this manufacturing scenario?

The Attempt at a Solution



partial derivatives part A)
[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP31819hf7fg47cd2027b00002b4b8bbei42a0a36?MSPStoreType=image/gif&s=34&w=196&h=36

[PLAIN]http://www4d.wolframalpha.com/Calculate/MSP/MSP62719hf7f989cef27630000665714d1f4f917d4?MSPStoreType=image/gif&s=35&w=273&h=42

Part B) substitute (50,25) into the partial derivative d/dy and explain what your answer means in the context of this manufacturing scenerio

[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP71819hf7fb0igibfd8h000068e7f2217hhi46eg?MSPStoreType=image/gif&s=7&w=269&h=43
[PLAIN]http://www4b.wolframalpha.com/Calculate/MSP/MSP72219hf7fb0igibfd8h00004b0cd1cei6c98c94?MSPStoreType=image/gif&s=7&w=156&h=36

I get a negative answer and I am not sure what this means in context of the manufacturing scenario.
 
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  • #2
P=-13.93 would simply mean that your profit is $-13.93, which is a $13.93 loss.
 
  • #3
so substituting in (50,25) means that 50 blackboard ($x) and 25 whiteboards ($y) were sold and it was a monthly loss of 13.93$ ??
 

FAQ: Partial Derivatives used in a manufacturing scenario word problem

What are partial derivatives and how are they used in manufacturing?

Partial derivatives are mathematical tools used to calculate how a specific variable affects the output of a function. In manufacturing, they are used to determine how changes in one variable, such as the production rate, will impact the overall production process.

Can you give an example of a manufacturing scenario where partial derivatives are used?

One example could be a company that produces smartphones. They may use partial derivatives to determine how changes in the production rate of a specific component, such as the screen, will affect the overall production cost and efficiency.

How do partial derivatives help in optimizing manufacturing processes?

Partial derivatives allow manufacturers to understand the relationship between different variables in their production process and optimize them accordingly. By calculating the impact of changes in each variable, they can make informed decisions to improve efficiency and reduce costs.

What other applications do partial derivatives have in manufacturing?

Aside from optimizing production processes, partial derivatives can also be used in quality control. By analyzing the derivative of a function, manufacturers can identify the optimal range of variables that will produce the highest quality products.

Are there any limitations to using partial derivatives in manufacturing?

While partial derivatives are a powerful tool, they may not accurately represent real-world scenarios in manufacturing. This is because they assume that all variables are continuous and change independently, which may not always be the case in a complex production process.

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